When N_2 goes to N_2^+, the N-N bond distance | vedclass

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(B) According to Molecular Orbital Theory, bond order is given by $\frac{1}{2}(N_b - N_a)$.For $N_2$ ($14$ electrons), the bond order is $3$. When it

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Increase, decrease

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(B) According to Molecular Orbital Theory, bond order is given by $\frac{1}{2}(N_b - N_a)$.For $N_2$ ($14$ electrons), the bond order is $3$. When it forms $N_2^+$, an electron is removed from a bonding molecular orbital, so the bond order decreases to $2.5$. Since bond distance is inversely proportional to bond order, the $N-N$ bond distance increases.For $O_2$ ($16$ electrons), the bond order is $2$. When it forms $O_2^+$, an electron is removed from an antibonding molecular orbital ($\pi^*$), so the bond order increases to $2.5$. Consequently, the $O-O$ bond distance decreases.

When $N_2$ goes to $N_2^+$, the $N-N$ bond distance ..... and when $O_2$ goes to $O_2^+$, the $O-O$ bond distance .......

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